Solution:
Total number of outcomes = 5 × 4 × 3 × 2 = 120
he number of favourable cases = 2 (4 × 3 × 2) - = 48 (i.e., odd numbers)
herefore,Required probability
$=\frac{48}{120}=\frac{2}{5}.$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column-$I$ | Column-$II$ |
| $(A)$ In a triangle $\triangle X Y Z$, let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$, respectively. If $2\left(a^2-b^2\right)=c^2$ and $\lambda=\frac{\sin (X-Y)}{\sin Z}$, then possible values of $n$ for which $\cos ( n \pi \lambda)=0$ is (are) | $(P)$ $1$ |
| $(B)$ In a triangle $\triangle X Y Z$, let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$, respectively. If $1+\cos 2 X-$ $2 \cos 2 Y=2 \sin X \sin Y$, then possible value(s) of $\frac{a}{b}$ is (are) | $(Q)$ $2$ |
| $(C)$ In $R ^2$, let $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $\beta \hat{i}+(1-\beta) \hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect of the origin $O$, respectively. If the distance of $Z$ from the bisector of the acute angle of $\overline{O X}$ with $\overline{O Y}$ is $\frac{3}{\sqrt{2}}$, then possible value(s) of $|\beta|$ is (are) | $(R)$ $3$ |
| $(D)$ Suppose that $F(\alpha)$ denotes the area of the region bounded by $x=$ $0, x=2, y^2=4 x$ and $y=|\alpha x-1|+|\alpha x-2|+\alpha x$, where $\alpha \in\{0$, 1\}. Then the value(s) of $F(\alpha)+\frac{8}{3} \sqrt{2}$, when $\alpha=0$ and $\alpha=1$, is (are) | $(S)$ $5$ |
| $(T)$ $6$ |
$f(x)=e^{x-1}-e^{-|x-1|} \text { and } g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right) \text {. }$ Then the area of the region in the first quadrant bounded by the curves $y=f(x), y=g(x)$ and $x=0$ is